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• Level: GCSE
• Subject: Maths
• Word count: 3461

# Beyond Pythagoras - Year 10 Maths Coursework

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Introduction

Beyond Pythagoras - Year 10 Maths Coursework

The numbers 3, 4 and 5 satisfy the condition 3²+4²=5²,

Because 3²= 3x3 =9
4²= 4x4 =16
5²= 5x5 =25

And so… 3²+4² (=9+16) = 25 (=5²)

I now have to find out if the following sets of numbers satisfy a similar condition of:

Smallest number² + Middle number² = Largest number²
(S² + M = L²)

a) 5, 12, 13

5² + 12² = 13²        ~   (25 + 144 = 169)

b) 7, 24, 25

7²+24² = 25²                ~   (49 + 576 = 625)

Here is a table containing the results:

 Short Side Middle Side Long Side 3 4 5 5 12 1 7 24 25

I looked at this table and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side.

I already know that the Smallest number² + Middle number² = Largest number². So I know that there will be a connection between the numbers written above. The only problem is that it is
obviously not:

Middle Side² + Large Side² = Small Side²

Because, 12² + 13² = 144 + 169 = 313
and 5² = 25

The difference between 25 and 313 is 288 which is far to big, so this means that the equation I want has nothing to do with 3 sides squared.

I will now try 2 sides squared.

M² + L = S²
= 12² + 13 = 5²
= 144 + 13 = 25
= 157 = 25

This does not work and neither will 13², because it is larger than 12². There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side squared.

12² + 13 = 5

This couldn’t work because 12² is already larger than 5, this also goes for 13². The only number now I can try squaring is the smallest number.

12 + 13 = 5²
25 = 25

This works with 5 being the smallest number/side but I need to know if it works with the other 2 triangles I know.

4 +5 = 3²
9 = 9

And…

24 + 25 = 7²
49 = 49

It works with both of my other triangles. So…

M + L = S²

Middle

4n2 - 4(n - 1)² = Middle side.

This should in theory give me my middle side. I will test my theory with the first term.

4 x 12 - 4(1-1)2 = 4
4 x 1 - 4 x 02 = 4
4 - 4 x 0 = 4
4 - 0 = 4
4 = 4

My formula works for the first term. I will now check if it works using the 2nd term.

4 x 22 - 4(2-1)2 = 12
4 x 4 - 4 x 12 = 12
16 - 4 x 1 = 12
16 - 4 = 12
12 = 12

My formula also works for the 2nd term. It’s looking likely that this is the correct formula. Just to check, I will check if it works using the 3rd term.

4 x 32 - 4(3-1)2 = 24
4 x 4 - 4 x 22 = 24
36 - 4 x 4 = 24
36 – 16 = 24
20 = 24
My formula doesn’t work for the 3rd term. It now looks as if “4n² - 4(n - 1)2” is not the correct formula after all. To check, I will look to see if the formula works using the 4th term.

4 x 42 - 4(4-1)2 = 40
4 x 16 - 4 x 32 = 40
64 - 4 x 9 = 40
64 – 36 = 40
28 = 40

My formula doesn’t work for the 4th term either. I can now safely say that 4n² - 4(n-1)² is definitely not the correct formula for the middle side.
I believe the problem with 4n² - 4(n-1)2 was that 4n², once you start using larger numbers, becomes far to high to bring it back down to the number that I want for the middle side. Also, 4(n-1)² is not as small when it gets larger so it doesn’t bring the 4n² down enough, to equal the middle side.
I know that the final formula will have something to do with 4 and have to be n². I will now try n² + 4.

Conclusion

Perimeter - Shortest Side = Relationship.
4n² + 6n + 2 – 2n + 1 = Relationship.
4n² + 4n + 1 = Relationship.

I am certain that this is the write answer. So…

Relationship between the Perimeter and Shortest side = 4n² + 4n + 1

The area is even simpler because, all you have to do is knock the 2n + 1 out of the equation. So…

Relationship between the Area and Shortest side = 2n² +2n
2
Here is the final table.

 S M L P A 3 4 5 12 6 5 12 13 30 30 7 24 25 56 984 9 40 41 90 180 11 60 61 132 330 13 84 85 182 546 15 112 113 240 840 17 144 145 306 1224 19 180 181 380 1710 21 220 221 462 2310

I would like to know whether or not the Pythagorean triple 3, 4, 5 is the basis of all triples just some of them.

To find this out I have been to the library and looked at some A-level textbooks and also searched on the internet. I have learnt ‘Arithmetic Progression’

3, 4, 5 is a Pythagorean triple

The pattern is plus one

f a = 3 and d = difference (which is +1) then

3 = a

4 = a + d

5 = a +2d

a, a +d, a + 2d

Therefore if you incorporate this into Pythagoras theorem

a2 + (a + d)2 = (a + 2d)2

a2 + (a + d)(a + d) = (a + 2d)2

a2 + a2 + ad + ad + d2 = (a + 2d)2

2a2 + 2ad + d2 = (a + 2d)2

2a2 + 2ad + d2 = (a + 2d)(a + 2d)

If you equate these equations to 0 you get

a2 – 3d2 – 2ad = 0

Change a to x

x2 – 3d2 – 2dx = 0

Factorise this equation to get

(x + d)(x – 3d)

Therefore

x = -d

x = 3d

x = -d is impossible as you cannot have a negative dimension

a, a + d, a + 2d

Is the same as

3d, 4d, 5d

This tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.

This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.

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